Problem: The Olympic Games in Bercouver are in full swing now. Here everyone has their own objectives: sportsmen compete for medals, and sport commentators compete for more convenient positions to give a running commentary. Today the main sport events take place at three round stadiums, and the commentator's objective is to choose the best point of observation, that is to say the point from where all the three stadiums can be observed. As all the sport competitions are of the same importance, the stadiums should be observed at the same angle. If the number of points meeting the conditions is more than one, the point with the maximum angle of observation is prefered. 

Would you, please, help the famous Berland commentator G. Berniev to find the best point of observation. It should be noted, that the stadiums do not hide each other, the commentator can easily see one stadium through the other.

Input

The input data consists of three lines, each of them describes the position of one stadium. The lines have the format x, y, r, where (x, y) are the coordinates of the stadium's center ( - 103 ≤ x, y ≤ 103), and r (1 ≤ r ≤ 103) is its radius. All the numbers in the input data are integer, stadiums do not have common points, and their centers are not on the same line. 

Output

Print the coordinates of the required point with five digits after the decimal point. If there is no answer meeting the conditions, the program shouldn't print anything. The output data should be left blank.

Examples

Input

0 0 10
60 0 10
30 30 10


Output

30.00000 0.00000

Solution (in PYTHON3):

Solution:
"""
Codeforces
2C - Commentator problem
http://codeforces.com/problemset/problem/2/C

Héctor González Belver
../06/2018
"""
import sys
import math
import functools

EPS = 1e-4

def points_same_observation_angle(circle1, circle2):
  x1,y1,r1 = circle1
  x2,y2,r2 = circle2

  #Let R be the distance from point А to a circle with center O and radius r. 
  #From this point, the angle formed by two tangents to circle: 2arcsin(r/R).
  #Two circles are observed at the same angle if r1/R1 = r2/R2
  #We obtain a equation of second degree: ax^2 + ay^2 + bx + cy + d = 0
  #If a=0 (r1=r2) --> represents a line
  #If a<>0 --> represents a circle

  a = r2*r2 - r1*r1
  b = 2*x2*r1*r1 - 2*x1*r2*r2
  c = 2*y2*r1*r1 - 2*y1*r2*r2
  d = (x1*x1 + y1*y1)*r2*r2 - (x2*x2 + y2*y2)*r1*r1

  return a, b, c, d

def circle_center_radius_from_equation(a,b,c,d):
  #--General equation of a circle--
  #Circle with center (h,k) and radius r
  #(x-h)^2 + (y-k)^2 = r^2 ==> x^2 + y^2 + Ax + By + C = 0
  # with A=-2h, B=-2k, C=h^2+k^2-r^2
  #
  #If ax^2 + ay^2 + bx + cy + d = 0
  #then b/a=-2h, c/a=-2k, d/a = h^2+k^2-r^2

  ox = b/(-2*a)  
  oy = c/(-2*a)
  r = math.sqrt(ox*ox + oy*oy - d/a)  
 
  return ox, oy, r

def lines_intersection(line1, line2):
  #Linear system:
  #a1 * x + b1 * y = c1
  #a2 * x + b2 * y = c2

  a1, b1, c1 = line1
  a2, b2, c2 = line2

  #Cramer's rule
  D = a1 * b2 - a2 * b1
  Dx = c1 * b2 - c2 * b1
  Dy = a1 * c2 - a2 *c1

  if math.fabs(D) >= EPS:
     INTXN_x1 = Dx / D
     INTXN_y1 = Dy / D
     return {(INTXN_x1, INTXN_y1)}
  else:
    #Same line (infinite solutions) or parallel (no solution)
    return None

def line_circle_intersection(line1, circle1):
  #Solve for the variable x with the formulas: ax + by = c (equation of line) 
  #and (x-h)^2 + (y-k)^2 = r^2 (equation of circle where h,k and r are known) 
  #and expand to obtain quadratic equation:
  #Ax^2 + Bx + C = 0
   
  a1,b1,c1 = line1
  x1,y1,r1 = circle1

  A = 1 + a1*a1 / (b1*b1)
  B = -2 * x1 + 2*a1/b1 * (-c1/b1 + y1)
  C = x1*x1 + (c1/b1 - y1)**2 -r1*r1

  discriminant = B*B - 4*A*C

  if math.fabs(discriminant) < EPS:
    #line is a tangent
    INTXN_x1 = -B / (2*A)
    INTXN_y1 = (c1 - a1*INTXN_x1) / b1

    return {(INTXN_x1, INTXN_y1)}
  elif discriminant < 0:
    #No intersection
    return None
  else:
    discriminant = math.sqrt(discriminant)
    
    INTXN_x1 = (-B + discriminant) / (2*A)
    INTXN_y1 = (c1 - a1*INTXN_x1) / b1

    INTXN_x2 = (-B - discriminant) / (2*A)
    INTXN_y2 = (c1 - a1*INTXN_x2) / b1

    return {(INTXN_x1, INTXN_y1), (INTXN_x2, INTXN_y2)}

def circles_intersection(circle1, circle2):
  x1,y1,r1 = circle1
  x2,y2,r2 = circle2

  dx = x2 - x1
  dy = y2 - y1

  dist_centers = math.hypot(dx, dy)

  if dist_centers < EPS and math.fabs(r1 - r2) < EPS:
    #Circles are coincident (infinite solutions)
    return None
  elif dist_centers < math.fabs(r1 - r2):
    #One circle is contained within the other (no solutions)
    return None
  elif dist_centers > r1 + r2:
    #Separate circles (no solutions)
    return None 
  else:
    #Single intersection (dist_centers = abs(r1+-r2)) and two points intersection
    
    #point3(x3,y3) is the point where the line 
    #through the circle intersection points crosses the line between the circle centers--> point1(x1,y1) and point2(x2,y2)

    #Distance from point1 to point3  
    a = (r1*r1 - r2*r2 + dist_centers*dist_centers ) / (2*dist_centers)
    #Distance from point3 to either of the intersection points
    h = math.sqrt((r1*r1 - a*a))

    #Coordinates of point3
    x3 = x1 + (dx*a/dist_centers)
    y3 = y1 + (dy*a/dist_centers)

    #Offsets of the intersection points from point3
    offset_x = -dy * (h/dist_centers);
    offset_y = dx * (h/dist_centers)
 

    INTXN_x1 = x3 + offset_x
    INTXN_y1 = y3 + offset_y

    INTXN_x2 = x3 - offset_x
    INTXN_y2 = y3 - offset_y

    return {(INTXN_x1, INTXN_y1), (INTXN_x2, INTXN_y2)}
      
  

def main():
  circle1, circle2, circle3 = (tuple(map(float,sys.stdin.readline().strip().split())) for _ in range(3))

  a1, b1, c1, d1 = points_same_observation_angle(circle1,circle2)
  a2, b2, c2, d2 = points_same_observation_angle(circle1,circle3)

  if math.fabs(a1) < EPS and math.fabs(a2) < EPS:
    #two lines
    INTXN_line1 = (b1, c1, -d1)
    INTXN_line2 = (b2, c2, -d2)

    sol = lines_intersection(INTXN_line1, INTXN_line2)

  elif math.fabs(a1) < EPS or math.fabs(a2) < EPS:
    #one line one circle
    if math.fabs(a1) < EPS:
      INTXN_line1 = (b1, c1, -d1)
      INTXN_circle1 = circle_center_radius_from_equation(a2, b2, c2, d2)
    else:
      INTXN_line1 = (b2, c2, -d2)
      INTXN_circle1 = circle_center_radius_from_equation(a1, b1, c1, d1)
    
    sol = line_circle_intersection(INTXN_line1, INTXN_circle1)

  else:
    #two circles
    INTXN_circle1 = circle_center_radius_from_equation(a1, b1, c1, d1)
    INTXN_circle2 = circle_center_radius_from_equation(a2, b2, c2, d2)

    sol = circles_intersection(INTXN_circle1, INTXN_circle2)

  if sol is None:
    sys.stdout.write('\n')
  else:
    #the point with the maximum angle of observation is prefered (2arcsin(r/R))==> max(r/R)
    #Reference: circle1
    f = functools.partial(lambda p1,p2:p2[2]/math.hypot(p1[0]-p2[0],p1[1]-p2[0]),p2=circle1)
    best_sol = max(sol,key=f)
    sys.stdout.write('{0:.5f} {1:.5f}'.format(best_sol[0], best_sol[1]) + '\n')

if __name__ == '__main__': 
  main()